Techniques: Substitution & Integration by Parts

Integration is the reverse of differentiation — and it is harder

To find an antiderivative of f, you are asked: what function has f as its derivative? Differentiation has reliable rules that grind any expression down to its slope. Going the other way is more like guessing-with-structure — you recognize a shape and undo the rule that would have produced it. Because of that, integration is genuinely a craft, not a recipe. The good news: two reversal techniques cover an enormous share of the integrals you will ever meet, and both come straight from rules you already know.

u-substitution: the chain rule, backwards

The chain rule says d/dx f(g(x)) = f'(g(x)) g'(x): when you differentiate a nested function, an extra factor g'(x) falls out. u-substitution is the art of spotting that leftover factor inside an integral and folding the inside function back up. The move is to rename the inside as u, so that u = g(x) and du = g'(x) dx. If the integral really is f'(g(x)) g'(x) dx, those pieces line up perfectly and the messy integral in x becomes a clean one in u.

Goal: integral of 2x cos(x^2) dx. The tricky part is the inside x^2, and notice its derivative 2x is sitting right there as a factor. Let u = x^2.
Then du = 2x dx — exactly the rest of the integrand. Substitute: integral of 2x cos(x^2) dx = integral of cos(u) du.
Integrate in u: integral of cos(u) du = sin(u) + C. Now substitute x^2 back for u: the answer is sin(x^2) + C.
Check by differentiating: d/dx sin(x^2) = cos(x^2) times 2x — the original integrand. The chain rule confirms it.

Integration by parts: the product rule, backwards

When the integrand is a product of two unrelated functions — x times e^x, or x times ln(x) — substitution usually fails, because there is no inner function whose derivative is hiding in the rest. The fix comes from reversing the product rule. Since (uv)' = u'v + uv', integrating both sides and rearranging gives the integration by parts formula: integral of u dv = uv - integral of v du. You trade the integral you cannot do for a (hopefully easier) one, paying with the simple product term uv.

The whole game is choosing which factor is u (you will differentiate it) and which is dv (you will integrate it). A handy guide is LIATE: prefer u to be whichever factor comes first in Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential. The idea is that you want u to get simpler when differentiated, and dv to be something you can actually integrate. Pick u = ln(x) over u = x; pick u = x over u = e^x.

integral of x e^x dx     (Algebraic times Exponential -> u = x)
  u = x,      dv = e^x dx
  du = dx,    v  = e^x
  = uv - integral of v du
  = x e^x - integral of e^x dx
  = x e^x - e^x + C

x e^x: choose u = x (it differentiates to 1, getting simpler) and dv = e^x dx. The leftover integral of e^x is trivial.

integral of x ln(x) dx   (Logarithmic wins -> u = ln x)
  u = ln x,   dv = x dx
  du = dx/x,  v  = x^2 / 2
  = (x^2/2) ln x - integral of (x^2/2)(1/x) dx
  = (x^2/2) ln x - integral of (x/2) dx
  = (x^2/2) ln x - x^2/4 + C

x ln(x): here you must pick u = ln(x), because ln(x) has no elementary antiderivative on its own but differentiates neatly to 1/x.

An honest limit: not everything can be integrated in closed form

Here is something textbooks sometimes hide. Every continuous function does have an antiderivative — that is guaranteed by the Fundamental Theorem of Calculus. But for many perfectly ordinary functions, that antiderivative cannot be written using the elementary functions (polynomials, roots, exp, log, trig and their combinations). Classic examples are e^{-x^2} (the bell curve), sin(x)/x, and 1/ln(x). No amount of clever substitution or parts will ever close them, because no elementary formula exists — this is a proven fact, not a gap in your skill.